# [R] create a new dataframe with intervals and computing a weighted average for each of its rows

Bert Gunter gunter.berton at gene.com
Sun Nov 24 16:19:24 CET 2013

```This post is complete garbage, and a great example of why not
bothering to read or follow the posting guide will cause a post to be
ignored.

1. It was not posted in plain text as the posting guide asks.

2. dput() was not used to pass example data

3. It appears the OP has not done due diligence by going through the
Introduction to R or other online tutorials to learn how R works,
although the post was so garbled that I may be wrong about that. My
apology, if so.

Cheers,
Bert

On Sun, Nov 24, 2013 at 1:08 AM, Luis Miguel Cerchiaro Barros
<luis_cerchiaro at hotmail.com> wrote:
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> I need you help with this problem, I have a data-frame like this:
>     BHID=c(43,43,43,43,44,44,44,44,44)    FROM=c(50.9,46.7,44.2,43.1,52.3,51.9,49.3,46.2,42.38)    TO=c(46.7,44.2,43.1,40.9,51.9,49.3,46.2,42.38,36.3)    AR=c(45,46,0.0,38.45,50.05,22.9,0,25,9)    DF<-data.frame(BHID,FROM,TO,VALUE)        #add the length     DF\$LENGTH=DF\$FROM-DF\$TO
> where:
> + BHID: is the borehole identification+ FROM: is  the start for every interval+ TO: is the end for every interval+ AR: is the value of our variable+ LENGTH: is the distance between FROM and TO
> what I want, is create a data frame which is "normalized", it means that every interval has the same length and the column **AR** is calculated as a Weighted arithmetic mean from the old **AR** and  **LENGTH** as its weight.
> For more clarity I going to show you how should look the desire data frame.
>     BHID        FROM    TO          AR          LENGTH    43        50.9        47.9       45.0     3.0    43       47.9        44.9       45.6     3.0    43       44.9        41.9       26.113      3.0    43            41.9        40.9    38.45        1.0    44....
> where:
> 1. AR is the Weighted arithmetic mean
> I have to make a clarification about the result:
> here I attached an example of my excel table with calculations:
>     ROW_ID BHID NEW_FROM NEW_TO NEW_AR  OLD_FROM OLD_TO WEIGHTS OLD_AR    1                 43   50.9         47.9              45               50.9    46.7              3.0      45    2                 43   47.9         44.9              45.6             50.9    46.7              1.2      45    2                 43   47.9         44.9                               46.7    44.2              1.8      46    3                 43   44.9         41.9              26.113    46.7   44.2              0.7      46    3                 43   44.9         41.9                                44.2   43.1              1.1      0    3                  43   44.9         41.9                               43.1    40.9              1.2      38.45    4              43   41.9         40.9              38.45     43.1   40.9              1.0      38.45
>
> you see guys, the NEW_AR is the weighted mean of the OLD_AR and its weights are in the column WEIGHTS.
> If you see the column LENGTH in the original data frame you can see, that the values are different, with the "normalization" we try to make that LENGTH uniform, in this case we choose the value 3.0 of course the last value of each borehole data could had a different LENGTH in this case 1.0
> What I have done to achieve the result
> OK guys in first place I have to say, I am not a professional and I am still learning  how to use R,
> my approximation is not elegant, I am trying to take the start and end of each borehole and use the function skeleton what I wrote, to create an uniform skeleton for the whole dataframe.
>     skeleton<-function(DF,LEN){    # define function to create a new skeleton    divide.int<-function(FROM,TO,div){    n=as.integer((FROM-TO)/div)+1    from=seq(FROM,(FROM-(n-1)*div),-div)    to=seq(FROM-(n-(n-1))*div,FROM-(n-1)*div,-div)    to[n]=TO    range<-data.frame(BHID=borehole_names[i,1],FROM=from,TO=to) # create a data.frame class object    range<-range[!(range\$FROM==range\$TO),] # erase the last value    }    # subset the data set for every borehole    borehole_names<-unique(DF["BHID"]) # collars id with cores    borehole_number<-nrow(borehole_names)  # collar number    #define an empty data.frame     borehole_Out<-data.frame(BHID=integer(),FROM=numeric(),TO=numeric())    # initialize the counter    i=1    # from this point starts the loop---------------    while(i<=borehole_number){    DFi <- subset(DF, BHID %in% borehole_names[i,1]) # Individual data frame for each boreholes    # take the beginning and end of every BOREHOLE    startBH<-head(DFi\$FROM,1)    endBH<-t!
>  ail(DFi\$TO,1)    # create the normalized intervals    borehole_i<-divide.int(FROM=startBH,TO=endBH,div=LEN)    borehole_Out<-rbind(borehole_Out,borehole_i)    i=i+1    }    borehole_Out    }    # TEST------------------------------------------    TEST<-skeleton(DF=DF,LEN=3.0)    TEST\$LENGTH=TEST\$FROM-TEST\$TO
> later I am trying to use the packages PLYR or DATA.TABLE to calculate the weighted means in AR but as I said I just started to use R and don't understand yet how this packages work
> again thanks in advanced and sorry for my bumpy english
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> and provide commented, minimal, self-contained, reproducible code.

--

Bert Gunter
Genentech Nonclinical Biostatistics

(650) 467-7374

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